Question

我是Android新手并且一直在尝试使用Parse.com。他们网站上提供的代码片段非常有用。我已经能够将数据从我的应用程序发送到数据库，这意味着它正在连接到正确的数据库。但是我无法从中检索到正确的数据。无论我使用什么对象ID，它总是返回null或0（取决于它是字符串还是整数）。我的代码如下：

public class MainActivity extends ActionBarActivity {



    TextView tv;
    Button b;
    String playerName;
    ParseQuery<ParseObject> query = ParseQuery.getQuery("GameScore");
    //int score;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);

        setContentView(R.layout.activity_main);
        tv=(TextView) findViewById (R.id.tv);
        b=(Button) findViewById (R.id.button1);
        Parse.initialize(this, "n12BWwZwk*************3dyKnuIUU", "vVi4TwI***************4peRD1VwyJ9CHw");
        final ParseObject gameScore = new ParseObject("GameScore");
       /* gameScore.put("score", 900);
        gameScore.put("playerName", "Pranav");
        gameScore.put("cheatMode", false);
        gameScore.saveInBackground();
      */


        query.whereEqualTo("objectId", "PnfRaUtHjB");
        query.findInBackground(new FindCallback<ParseObject>() {

            @Override
            public void done(List<ParseObject> arg0, ParseException arg1) {
                // TODO Auto-generated method stub
                if (arg1==null)
                {
                    int score=gameScore.getInt("score");
                    tv.setText("Player score - "+score);
                    Log.d("PTesting", "Player Score");
                    //Toast.makeText(getApplicationContext(), "Reached till Toast "+score,
                        //     Toast.LENGTH_LONG).show();

                }
                else
                {
                    Log.d("score", "Error: " + arg1.getMessage());
                }
            }

        });

}
}

我已尝试将[ query.whereEqualto ]与playerName一起使用并获得分数。相同的结果。

我也试过这个：

ParseQuery<ParseObject> query=ParseQuery.getQuery("GameScore");
        query.getInBackground("YsHLfJ7tUB", new GetCallback<ParseObject>() {

            @Override
            public void done(ParseObject arg0, ParseException arg1) {
                // TODO Auto-generated method stub

                if (arg1==null)
                {
                    int score = gameScore.getInt("score");
                    playerName = gameScore.getString("playerName");
                    boolean cheatMode = gameScore.getBoolean("cheatMode");
                    tv.setText("Updating "+playerName);

                }
                else
                {
                    Log.d("score", "Error: " + arg1.getMessage());
                }
            }

返回的值为0或null。

数据库如下所示：

http://i.imgur.com/SsUO8AG.jpg

Answer 1

您正在使用错误的对象进行检索。试试这个。

ParseQuery<ParseObject> query=ParseQuery.getQuery("GameScore");
        query.getInBackground("YsHLfJ7tUB", new GetCallback<ParseObject>() {

            @Override
            public void done(ParseObject arg0, ParseException arg1) {
                // TODO Auto-generated method stub

                if (arg1==null)
                {
                    int score = arg0.getInt("score");
                    playerName = arg0.getString("playerName");
                    boolean cheatMode = arg0.getBoolean("cheatMode");
                    tv.setText("Updating "+playerName);

                }
                else
                {
                    Log.d("score", "Error: " + arg1.getMessage());
                }
            }

我建议使用子类进行解析。 http://blog.parse.com/2013/05/30/parse-on-android-just-got-classier/

如何从parse.com数据库（Android）中检索数据？

1 个答案: